ASSIGNMENT 5. IS 605 FUNDAMENTALS OF COMPUTATIONAL MATHEMATICS

Choose independently two numbers B and C at random from the interval [0, 1] with uniform density. Prove that B and C are proper probability distributions. Note that the point (B,C) is then chosen at random in the unit square.

Solution:

B and C are chosen randomly from the interval of unifrom density, so they should create uniform distribution with density 1/[0,1] or 1. Let’s test it.

# Clear plots
if(!is.null(dev.list())) dev.off()

## null device 
##           1

# Clear console
cat("\014") 



# Clean workspace
rm(list=ls())

B<-runif(100000,0,1) 

C<-runif(100000,0,1) 

hist(B)

hist(C)

The histograms look uniform

Find the probability that

1

1. B + C < 1/2.

Mathematical solution:

Please note that solving B+C=1/2. We get C=1/2-B

C1 <- function(B1) {0.5-B1}

# because C cannot be negative B will be from 0 to 0.5

integrate(C1,lower=0, upper=0.5)

## 0.125 with absolute error < 1.4e-15

Analytical solution

prob<-0

x<-c(0.0)

y<-c(0.0)

max<-500

for (i in (1:max))
{for (j in (1:max))
  {  if (B[i]+C[j]<0.5) {prob<-prob+1/max^2
 x<-c(x,B[i])
  y<-c(y,C[j])
  }
}
}

prob

## [1] 0.115836

plot(x,y)

Answer is pretty close.

2

2. BC < 1/2.

C1 <- function(B1) {0.5/B1}

# because C cannot be more than 1 then for B from 0 to 0.5 it will be 1

myintegral<-integrate(C1,lower=0.5, upper=1)

1*0.5+myintegral$value

## [1] 0.8465736

Analytical solution

prob<-0

x<-c(0.0)

y<-c(0.0)

max<-300

for (i in (1:max))
{for (j in (1:max))
  {if (B[i]*C[j]<0.5)
    {prob<-prob+1/max^2
    x<-c(x,B[i])
  y<-c(y,C[j])
  }
}
}

prob

## [1] 0.8359444

plot(x,y)

Answer is pretty close.

3

3. |B - C| < 1/2.

# We have for B on interval [0,0.5)

C1 <- function(B1) {B1+0.5}

# And we have for B on interval [0.5,1] 

C2<-function(B1) {B1-0.5}

# We need to subtract integral of C2 from 1*0.5

integrate(C1,lower=0, upper=0.5)$value+(0.5-integrate(C2,lower=0.5, upper=1)$value)

## [1] 0.75

Analytical solution

prob<-0

x<-c(0.0)

y<-c(0.0)

max<-300

for (i in (1:max))
{for (j in (1:max))
  {
  if (abs(B[i]-C[j])<0.5) {prob<-prob+1/max^2
  x<-c(x,B[i])
  y<-c(y,C[j])
  }
}
}

prob

## [1] 0.7502444

plot(x,y)

Close

4

4. max{B,C} < 1/2

This one is easy to solve using analytical method

prob<-0

x<-c(0.0)

y<-c(0.0)

max<-400

for (i in (1:max))
{for (j in (1:max))
  {
  if (max(B[i],C[j])<0.5) {
  prob<-prob+1/max^2
  x<-c(x,B[i])
  y<-c(y,C[j])
  }
}
}

prob

## [1] 0.2374313

plot(x,y)

Answer is 0.25

5

5. min{B,C} < 1/2

This one is easy to solve using analytical method

prob<-0

x<-c(0.0)

y<-c(0.0)

max<-400

for (i in (1:max))
{for (j in (1:max))
  {
  if (min(B[i],C[j])<0.5) {
  prob<-prob+1/max^2
  x<-c(x,B[i])
  y<-c(y,C[j])
  }
}
}

prob

## [1] 0.7375687

plot(x,y)

Answer is 0.75